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I'm trying to prove this interesting result. Could you verify if my attempt is fine?

Let $(X_i, d_i)_{i=1}^n$ be a finite collection of metric spaces. Let $p \in [1, \infty]$ and $\| \cdot \|_p$ be the $p$-norm on $\mathbb R^n$. Consider $D_p$ defined on $X := X_1 \times \cdots \times X_n$ by $$ D_p ((x_1, \ldots, x_n), (y_1, \ldots, y_n)) := \|(d_1(x_1, y_1), \ldots , d_n (x_n, y_n))\|_p. $$

Then $D_p$ is a metric. All norms in a finite-dimensional vector space are equivalent, so all $D_p$ for $p \in [1 , \infty]$ are equivalent and thus induce the same topology $\tau$. Let $\tau_i$ be the topology on $X_i$ induced from metric $d_i$.

Theorem: $\tau$ coincides with the product topology $\tau'$ of $\tau_1, \ldots, \tau_n$.

I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.

Update: The case of countably infinite cartesian product of metric spaces is discuss here.

Analyst
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1 Answers1

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It suffices for us to consider $p=\infty$. In this case, $$ D_\infty ((x_1, \ldots, x_n), (y_1, \ldots, y_n)) = \max \{d_1(x_1, y_1), \ldots , d_n (x_n, y_n)\}. $$

It suffices to prove that $\tau$ and $\tau'$ have the same system of neighborhoods (nbh) at each point. Fix $x \in X$.

Let $U$ be a nbh of $x$ in $\tau$. There is $r \in \mathbb R_{>0}$ such that $B(x, r) := \{y \in X \mid D_\infty(y, x) < r\} \subset U$. Then $$ \begin{align} B(x, r) &= \{y \in X \mid \forall i =1, \ldots, n:d_i(y_i, x_i) < r\} \\ &= \bigcap_{i=1}^n \{y \in X \mid d_i(y_i, x_i) < r\}. \end{align} $$

Notice that the map $y \mapsto d_i(y_i, x_i)$ is continuous w.r.t. $\tau'$, so $\{y \in X \mid d_i(y_i, x_i) < r\} \in \tau'$ for all $i =1, \ldots, n$. Hence $B(x, r) \in \tau'$ and thus $U$ is a nbh of $x$ in $\tau'$.

Let $U$ be a nbh of $x$ in $\tau'$. Then there is nbh $V_i$ of $x_i$ in $\tau_i$ such that $$ V_1 \times \cdots \times V_n \subset U. $$

Then there is $r_i \in \mathbb R_{>0}$ such that $B(x_i, r_i) := \{y\in X_i \mid d(y, x_i) < r_i\} \subset V_i$. Let $r := \min \{r_1, \ldots, r_n\}$. Then $V := \{y \in X \mid D_\infty(y, x) < r\} \subset U$. Clearly, $V \in \tau$. Then $U$ is a nbh of $x$ in $\tau$. This completes the proof.

Analyst
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